The union of non-disjoint connected is connected

Suppose are connected sets in and . Show that is connected. Does this hold true for ?

Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if are open in connected, and , then . Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.

Suppose are connected sets in and . Show that is connected. Does this hold true for ?

Okay, so I have seen proofs that rely on topological properties that we have not really discussed or proven in my class (like for instance, if are open in connected, and , then . Assuming that, I understand the proof for this theorem pretty well, but how do I prove that assumption? Or is there another way to solve this question without invoking open sets/topology in general? Thanks.

Since "being connected" is defined in terms of open sets, I don't think there is a way around talking of open sets and the like in a proof of the above proposition.

To show that the union of connected sets is connected, if , it suffices to show that given any two open sets with , and , it follows that either or .

And why is that? Well, suppose that , then we have either or (but not both, because U and V are disjoint).
If , then, because are both connected, it follows that and, and therefore .
Similarly, if it follows that .

Finally, to the question as to whether is connected if are connected and . The answer is no, for consider the intersection of a line and a circle in that consists of two isolated points. In such a case, the intersection is clearly not connected.

There is another characterization of connect sets. A set is connected if it is not the union of two separated setsTwo sets are separated if each is non-empty and neither contains a point nor a limit point of the other.

Suppose that are separated sets such .
We know that so .
In either case, because each of is connected there follows a contradiction.
So must be connected.